Finitely Generated Abelian Groups

[Combinatorics]

Homework Statement

Let $$p$$ be prime and let $$b_1 ,...,b_k$$ be non-negative integers. Show that if :
$$G \simeq (\mathbb{Z} / p )^{b_1} \oplus ... \oplus (\mathbb{Z} / p^k ) ^ {b_k}$$
then the integers $$b_i$$ are uniquely determined by G . (Hint: consider the kernel of the homomorphism $$f_i :G \to G$$ that is multiplication by $$p^i$$ . Show that $$f_1 , f_2$$ determine $$b_1$$.
Proceed similarly )

Homework Equations

The question, together with the theorem it should help me prove is attached (this question should help me prove the uniqueness part of the theorem)

The Attempt at a Solution

I've tried using the hint, and considering the homomorphism $$f_1 : G \to G$$ that is defined by $$f_1 (x ) = px$$ . Its kernel is $$ker f_1 = (\mathbb{Z}_p ) ^{b_1} \bigoplus (\mathbb{Z}_p) ^ {b_2} ...$$.
But how does it help me? Am I right in my calculation of the kernel? Hope someone will be able to help me

Thanks in advance !

 

[morphism]

The hint is essentially leading you to consider a proof by induction on k. Notice that $$ \ker f_1 = (\mathbb Z/p)^{b_1} \oplus (\mathbb Z/p)^{b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots = (\mathbb Z/p)^{b_1+b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots. $$

 

[Combinatorics]

The hint is essentially leading you to consider a proof by induction on k. Notice that $$ \ker f_1 = (\mathbb Z/p)^{b_1} \oplus (\mathbb Z/p)^{b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots = (\mathbb Z/p)^{b_1+b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots. $$

Can you please explain how did you get that this is the kernel?

Thanks a lot !

 

[morphism]

Well, if px=0 mod p^i, then p^i divides px, hence p^{i-1} divides x, i.e. x=0 mod p^i.